Hi, Neil and Seqfan
I think A082010 is not correct. it and the sequence of definition are different.
If you started from 0 then it must be the following.
0,1,2,1,2,....
It is not so interesting.
The first interesting case is the following.
3,5,9,15,25,41,66,33,53,85,137,220,110,55,89,143,229,367,588,294,147,236,118,59,95,153,245,393
http://list.seqfan.eu/pipermail/seqfan/2009-September/009818.html
Ask Don Reble.
He has 2000000 terms of the sequence.
Once Franklin wrote
>This can be simplified: if x is odd, then next x is floor(8/5*x)+1.
>No need for separate m and n; let r be the ratio (m/n), and if x is
>odd, the next x is floor(r*x)+1. And now we don't even need r to be
>rational.
You are right.
8/5 sequence is one of the K-Sequence.
a(n)=[A*a(n-1)+B]/p^m
Where [N] is integer part of N, p^m is the highest p power dividing [A*a(n-1)+B]
http://boat.zero.ad.jp/~zbi74583/ess0.htm
3 examples :
See A028948, A029580, A036982
>When r is the golden ratio, (sqrt(5)+1)/2, starting with 3, we get
>A001595, a sequence of all odd numbers, so there is unlimited growth.
>This may shed some light on what is happening with 8/5.
I think your opinion is interesting.
My observation is the following.
p=2, a(0)=3, A=1.6, 1<=B<2
In this area all sequences are unlimited.
You say the case of A=(1+root(5))/2 is also unlimited.
How did you find this fact?
I think it is your discovery.
Once I conjectured as follows.
"K-Sequence which represents Fibonacci Sequence exists"
I knew that A001959 is one of the K-Sequence.
So my conjecture is almost right.
Yasutoshi